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POWhy can't I convert 'char**' to a 'const char* const*' in C?
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copied!<p>The following code snippet (correctly) gives a warning in C and an error in C++ (using gcc & g++ respectively, tested with versions 3.4.5 and 4.2.1; MSVC does not seem to care):</p> <pre><code>char **a; const char** b = a; </code></pre> <p>I can understand and accept this.<br> The C++ solution to this problem is to change b to be a const char * const *, which disallows reassignment of the pointers and prevents you from circumventing const-correctness (<a href="http://www.parashift.com/c++-faq-lite/const-correctness.html#faq-18.17" rel="noreferrer">C++ FAQ</a>).<br></p> <pre><code>char **a; const char* const* b = a; </code></pre> <p>However, in pure C, the corrected version (using const char * const *) still gives a warning, and I don't understand why. Is there a way to get around this without using a cast?</p> <p>To clarify:<br> 1) Why does this generate a warning in C? It should be entirely const-safe, and the C++ compiler seems to recognize it as such.<br> 2) What is the correct way to go about accepting this char** as a parameter while saying (and having the compiler enforce) that I will not be modifying the characters it points to? For example, if I wanted to write a function:</p> <pre><code>void f(const char* const* in) { // Only reads the data from in, does not write to it } </code></pre> <p>And I wanted to invoke it on a char**, what would be the correct type for the parameter?</p> <p>Edit: Thank you to those who have responded, particularly those who addressed the question and/or followed up on my responses.</p> <p>I've accepted the answer that what I want to do cannot be done without a cast, regardless of whether or not it should be possible.</p>

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